R - 迭代地应用变量列表的函数

Question

我的目标是创建一个函数，当循环遍历数据框的多个变量时，将返回一个新数据框，其中包含每个变量每个级别的百分比和 95% 置信区间。

例如，如果我将此函数应用于 mtcars 数据框中的“cyl”和“am”，我希望将其作为最终结果：

  variable level                ci.95
1      cyl     4 34.38 (19.50, 53.11)
2      cyl     6 21.88 (10.35, 40.45)
3      cyl     8 43.75 (27.10, 61.94)
4       am     0  59.38 (40.94, 75.5)
5       am     1 40.62 (24.50, 59.06) 

所以，到目前为止，我的函数似乎适用于单个变量；但是，我有两个问题希望社区可以帮助我：

1. 常规 R-ifying 我的代码。我仍然是 R 新手。我已经阅读了足够多的帖子，知道 R 爱好者通常不鼓励使用 for 循环，但我仍然很难使用 apply 函数（在大多数情况下，这似乎是 for 循环的替代方法）。

2. 将此函数应用于变量列表 - 生成单个数据框，其中包含每个变量的每个级别的函数返回值。

到目前为止，我的代码如下所示：

t1.props <- function(x, data = NULL) {

  # Grab dataframe and/or variable name
  if(!missing(data)){
    var <- data[,deparse(substitute(x))]
  } else {
    var <- x
  }

  # Grab variable name for use in ouput
  var.name <- substitute(x)

  # Omit observations with missing data
  var.clean <- na.omit(var)

  # Number of nonmissing observations
  n <- length(var.clean)

  # Grab levels of variable
  levels <- sort(unique(var.clean))

  # Create an empty data frame to store values
  out <- data.frame(variable = NA,
                    level = NA,
                    ci.95 = NA)

  # Estimate prop, se, and ci for each level of the variable
  for(i in seq_along(levels)) {
    prop <- paste0("prop", i)
    se <- paste0("se", i)
    log.prop <- paste0("log.trans", i)
    log.se <- paste0("log.se", i)
    log.l <- paste0("log.l", i)
    log.u <- paste0("log.u", i)
    lcl <- paste0("lcl", i)
    ucl <- paste0("ucl", i)

    # Find the proportion for each level of the variable
    assign(prop, sum(var.clean == levels[i]) / n)

    # Find the standard error for each level of the variable
    assign(se, sd(var.clean == levels[i]) /
             sqrt(length(var.clean == levels[i])))

    # Perform a logit transformation of the original percentage estimate
    assign(log.prop, log(get(prop)) - log(1 - get(prop)))

    # Transform the standard error of the percentage to a standard error of its
    # logit transformation
    assign(log.se, get(se) / (get(prop) * (1 - get(prop))))

    # Calculate the lower and upper confidence bounds of the logit
    # transformation
    assign(log.l,
           get(log.prop) -
           qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
    assign(log.u,
           get(log.prop) +
           qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))

    # Finally, perform inverse logit transformations to get the confidence bounds
    assign(lcl, exp(get(log.l)) / (1 + exp(get(log.l))))
    assign(ucl, exp(get(log.u)) / (1 + exp(get(log.u))))

    # Create a combined 95% CI variable for easy copy/paste into Word tables
    ci.95 <- paste0(round(get(prop) * 100, 2), " ",
                "(", sprintf("%.2f", round(get(lcl) * 100, 2)), ",", " ",
                round(get(ucl) * 100, 2), ")")

    # Populate the "out" data frame with values
    out <- rbind(out, c(as.character(var.name), levels[i], ci.95))
  }

  # Remove first (empty) row from out
  # But only in the first iteration
  if (is.na(out[1,1])) {
    out <- out[-1, ]
    rownames(out) <- 1:nrow(out)
  }
  out
}

data(mtcars)
t1.props(cyl, mtcars)

感谢您提供的任何帮助或建议。

Answer 1

您使用的所有函数的好处是它们已经矢量化（sd 和 qt 除外，但您可以使用 Vectorize 轻松矢量化它们以获得特定参数）。这意味着您可以将向量传递给它们，而无需编写单个循环。我省略了处理准备输入和修饰输出的函数部分。

t1.props <- function(var, data=mtcars) {
    N <- nrow(data)
    levels <- names(table(data[,var]))
    count <- unclass(table(data[,var]))        # counts
    prop <- count / N                          # proportions
    se <- sqrt(prop * (1-prop)/(N-1))          # standard errors of props.
    lprop <- log(prop) - log(1-prop)           # logged prop
    lse <- se / (prop*(1-prop))                # logged se
    stat <- Vectorize(qt, "df")(0.975, N-1)    # tstats
    llower <- lprop - stat*lse                 # log lower 
    lupper <- lprop + stat*lse                 # log upper
    lower <- exp(llower) / (1 + exp(llower))   # lower ci
    upper <- exp(lupper) / (1 + exp(lupper))   # upper ci

    data.frame(variable=var,
               level=levels,
               perc=100*prop,
               lower=100*lower,
               upper=100*upper)
}

因此，当您将函数应用于多个变量时，唯一的显式应用/循环出现，如下所示

## Apply your function to two variables
do.call(rbind, lapply(c("cyl", "am"), t1.props))
#   variable level   perc    lower    upper
# 4      cyl     4 34.375 19.49961 53.11130
# 6      cyl     6 21.875 10.34883 40.44691
# 8      cyl     8 43.750 27.09672 61.94211
# 0       am     0 59.375 40.94225 75.49765
# 1       am     1 40.625 24.50235 59.05775

就代码中的循环而言，它在效率方面并不是特别重要，但是您可以看到代码简洁时可以更容易阅读 - 并且应用函数提供了很多简单的一个 -线解决方案。

我认为更改代码最重要的是使用assign 和get。相反，您可以将变量存储在列表或其他数据结构中，并在需要时使用setNames、names<- 或names(...) <- 来命名组件。

Answer 2

您也可以保持该功能基本不变并在其上使用lapply：

vars <- c("cyl", "am")
lapply(vars, t1.props, data=mtcars)
[[1]]
  variable level                ci.95
1      cyl     4 34.38 (19.50, 53.11)
2      cyl     6 21.88 (10.35, 40.45)
3      cyl     8 43.75 (27.10, 61.94)

[[2]]
  variable level                ci.95
1       am     0  59.38 (40.94, 75.5)
2       am     1 40.62 (24.50, 59.06)

并将它们全部合并到一个数据框中：

lst <- lapply(vars, t1.props, data=mtcars)
do.call(rbind,lst)

数据

您必须将var 和var.name 分配简化为：

t1.props <- function(x, data = NULL) {

  # Grab dataframe and/or variable name
  if(!missing(data)){
    var <- data[,x]
  } else {
    var <- x
  }

  # Grab variable name for use in ouput
  var.name <- x

  # Omit observations with missing data
  var.clean <- na.omit(var)

  # Number of nonmissing observations
  n <- length(var.clean)

  # Grab levels of variable
  levels <- sort(unique(var.clean))

  # Create an empty data frame to store values
  out <- data.frame(variable = NA,
                    level = NA,
                    ci.95 = NA)

  # Estimate prop, se, and ci for each level of the variable
  for(i in seq_along(levels)) {
    prop <- paste0("prop", i)
    se <- paste0("se", i)
    log.prop <- paste0("log.trans", i)
    log.se <- paste0("log.se", i)
    log.l <- paste0("log.l", i)
    log.u <- paste0("log.u", i)
    lcl <- paste0("lcl", i)
    ucl <- paste0("ucl", i)

    # Find the proportion for each level of the variable
    assign(prop, sum(var.clean == levels[i]) / n)

    # Find the standard error for each level of the variable
    assign(se, sd(var.clean == levels[i]) /
             sqrt(length(var.clean == levels[i])))

    # Perform a logit transformation of the original percentage estimate
    assign(log.prop, log(get(prop)) - log(1 - get(prop)))

    # Transform the standard error of the percentage to a standard error of its
    # logit transformation
    assign(log.se, get(se) / (get(prop) * (1 - get(prop))))

    # Calculate the lower and upper confidence bounds of the logit
    # transformation
    assign(log.l,
           get(log.prop) -
             qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
    assign(log.u,
           get(log.prop) +
             qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))

    # Finally, perform inverse logit transformations to get the confidence bounds
    assign(lcl, exp(get(log.l)) / (1 + exp(get(log.l))))
    assign(ucl, exp(get(log.u)) / (1 + exp(get(log.u))))

    # Create a combined 95% CI variable for easy copy/paste into Word tables
    ci.95 <- paste0(round(get(prop) * 100, 2), " ",
                    "(", sprintf("%.2f", round(get(lcl) * 100, 2)), ",", " ",
                    round(get(ucl) * 100, 2), ")")

    # Populate the "out" data frame with values
    out <- rbind(out, c(as.character(var.name), levels[i], ci.95))
  }

  # Remove first (empty) row from out
  # But only in the first iteration
  if (is.na(out[1,1])) {
    out <- out[-1, ]
    rownames(out) <- 1:nrow(out)
  }
  out
}